Question

**3a. Write the balanced reaction and solubility product
expression (K****SP****) for dissolving
silver chromate:
Ag****2****CrO****4(s)****.
Include all charges, stoichiometric coefficients, and phase
subscripts. (4 pts)**

**c. Determine [Ag****+1****]
and [CrO****4****−2****] fo
a saturated solution of
Ag****2****CrO****4**
**if K****SP** **= 1.1 ×
10****−12****. Also, determine the molar
solubility. (8 pts)**

**d. Use the K****SP** **above
to determine
[CrO****4****−2****] for a
solution that is saturated with
Ag****2****CrO****4****,
and also has a common ion effect where
[Ag****+1****] = 0.495 M. (4
pts)**

Answer #1

(a)

Solubility equation is,

**Ag2CrO4 (s)
2 Ag^+ (aq) + CrO4^2- (aq)**

Solubility product expression can be written as,

**Ksp = [Ag^+]^2[CrO4^2-]**

(b)

If solubility of Ag2CrO4 is S M, then [Ag^+] = 2 S M and [CrO4^2-] = S M

Ksp = S^2 x S

Ksp = S^3

S = (Ksp)^(1/3)

S = (1.1 x 10^-12)^(1/3)

**S = Solubility of Ag2CrO4 = 1.03 x 10^-4 M**

From the solubility equation,

**[Ag^+] = 2 x [Ag2CrO4] = 2 x 1.03 x 10^-4 = 2.06 x 10^-4
M**

**[CrO4^2-] = [Ag2CrO4] = 1.03 x 10^-4 M**

(c)

Solubility product expression can be written as,

Ksp = [Ag^+]^2[CrO4^2-]

1.1 x 10^-12 = (0.495)^2[CrO4^2-]

1.1 x 10^-12 = 0.245 [CrO4^2-]

[CrO4^2-] = 1.1 x 10^-12 / 0.245

**[CrO4^2-] = 4.08 x 10^-12 M**

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